text:
Scottish republicans urged to stay away from Orange Order parade
Activists advised to steer clear of Protestant march through Edinburgh amid fears of clashes damaging the yes campaign
Members of the Orange Lodges march through Glasgow city centre. Photograph: PA
Henry McDonald and Severin Carrell
Wed 3 Sep 2014 03.00 EDT
Scottish republicans have urged leftwing activists to ignore an anti-independence march by the Protestant Orange Order, which takes place five days before the referendum.
The James Connolly Society (JCS), a pro-Irish republican group which supports Scottish independence, has warned its supporters not to stage any protests against the Orange Order parade through central Edinburgh to avoid violent clashes damaging the yes campaign.
Jim Slaven, chairman of the JCS, one of Scotland's largest republican groups, said: "We are asking our members and supporters to concentrate on making the positive case for the break up of the UK state and not to respond to the Orange Order's provocation in coming weeks."
The parade on Saturday 13 September is expected to draw up to 15,000 members of Scottish and Northern Irish Orange Order lodges and flute bands, as well as the Loyalist Apprentice Boys of Derry, in one of the largest but most controversial pro-union rallies of the referendum campaign.
In parallel with Slaven's concerns, leaders of the official pro-UK campaign, Better Together, fear the parade by the militantly unionist order will alienate centre-left and Catholic no voters at a critical time and increase sectarian tensions.
The Orange Order parade comes only a day after Ukip leader Nigel Farage stages a party rally against independence in Glasgow.
Worried about the backlash boosting the yes vote, Better Together has refused to work with either the Orange Order or Ukip because they are regarded as too divisive. With the latest polls showing a narrow six-point gap between yes and no, their rallies in the last week of the campaign will unnerve no campaigners.
Slaven said the decision by City of Edinburgh council to allow the Grand Orange Lodge of Scotland to march was a disgrace, but added that republicans and leftwing groups should protest against the demonstration at a later date through official channels.
"The nature and timing of our opposition will be decided by republicans not by the Orange Order. These marches are a political problem and require a political solution," he said.
Slaven's group is named after the Edinburgh-born Marxist and Irish republican leader James Connolly, who was executed by the British army for taking part in the 1916 Easter rising in Dublin. It backs Scottish independence but opposes the Scottish National party's policies on retaining the Queen as head of state and using sterling.
This year Sam "Skelly" McCrory, the Loyalist commander of the Ulster Defence Association's prisoners in the Maze jail during the last years of the Troubles, had urged Northern Irish loyalists and Orange Order members to stay away from Scotland to avoid damaging the no campaign.
McCrory told the Guardian he feared Northern Irish loyalists would alienate Catholic voters and other centre-left Scots. "The case for the union over here is economic not emotional: it's about jobs, it's about the currency; it's not really about flying flags. Loyalists from Northern Ireland should leave Ulster's politics out of this," he said.
Money, pride and dirty tricks: what Scotland can learn from Quebec
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Scottish independence: momentum is with yes but country is divided
Jubilant yes voters greet SNP's Salmond and Sturgeon in Glasgow
Scottish independence referendum: two weeks to go – as it happened
David Cameron: I will not resign if Scotland votes for independence
Alex Salmond's vision for a fairer Scotland is a con, says Ed Miliband
Scottish referendum campaign 'permeated by something ugly' – cleric


There are multiple artists that go by the name of Lauren. (1) Lauren was a contestant on Australian Pop Idol. (2) Lauren is a male French singer who duetted with fellow French singer Kareen Antonn in the single, 'Il suffirait d'?tre un homme '.
(1) Lauren was a contestant on Australian Pop Idol.
(2) Lauren is a male French singer who duetted with fellow French singer Kareen Antonn in the single, 'Il suffirait d'?tre un homme '.


Blue Fin is located in Port MacDonnell part of the beautiful Limestone Coast Region.
It is a terrific house, with complimentary Wi-Fi, built for long lasting holiday memories. Plenty of room for the kids to kick the footy, ride their bikes or fly a kite or run without a care in the world. For the Adults it is quiet enough to read a book/magazine or drink a couple of glasses of wine (or 2) while solving the problems of the world.
Blue Fin is a new home with ducted reverse cycle air conditioning through out and a great back deck for sitting back with fresh seafood and wine.
Two LCD televisions in bedroom 1,3 and a large plasma in living room with stereo and ipod jack. Nintendo Wii, games, DVDs and plenty of books & magazines. A coffee pod machine and many other appliances make this a gourmet kitchen.
Very sheltered area to relax and unwind. Plenty of parking for the largest of boats and cars. With rainwater and power for flushing motors and boats, with the bonus of a sink at the rear of the shed for cleaning your daily prized catches.
Base Rate for 4 People then extra $20 per night per adult and $15 per night for children.


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Please enjoy browsing the materials.
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Guest Post by Prof. Contreras – CSIRO v. Cisco: The Convergence of RAND and non-RAND Royalties for Standards-Essential Patents
December 7, 2015 PatentDamages, Enablement, LicensesJason Rantanen
Guest Post by Jorge L. Contreras, Associate Professor, University of Utah College of Law.
In Commonwealth Scientific and Industrial Research Organisation v. Cisco Systems, Inc. (Fed. Cir., Dec. 1, 2015), the Federal Circuit established important new guidelines for the calculation of "reasonable royalty" damages for standards-essential patents (SEPs), even in the absence of the patent holder's commitment to license on reasonable and nondiscriminatory (RAND) terms. Chief Judge Prost, writing for a panel that also included Judges Dyk and Hughes, found that Chief Judge Leonard Davis of the Eastern District of Texas erred by failing, among other things, to account for the "standard-essential status" of a Commonwealth Scientific (CSIRO) patent infringed by Cisco. The decision signals another important step toward the convergence of "reasonable royalty" damages in RAND and other patent cases.
CSIRO is a leading Australian governmental research organization. In 1996 CSIRO obtained U.S. Patent 5,487,069, claiming techniques for addressing "multipath" problems in wireless signal processing. These techniques were later incorporated into IEEE's 802.11a ("Wi-Fi") standard, first published in 1999. In connection with the approval of 802.11a, IEEE requested, and CSIRO provided, a Letter of Assurance under which CSIRO committed to license the '069 patent to manufacturers of 802.11a-compliant products on "reasonable and nondiscriminatory" (RAND) terms. When later versions of 802.11 were developed, IEEE again requested that CSIRO commit to license the '069 patent on RAND terms. CSIRO, however, refused to issue such additional assurances.
In 2001, Cisco acquired Radiata, Inc., a company founded by a former CSIRO scientist to manufacture wireless chips. Radiata had entered into a Technology License Agreement (TLA) with CSIRO in 1998, under which Radiata paid CSIRO royalties for use of the '069 patent based on a percentage of Radiata's chip sale prices. These royalties ranged from 1% to 5% of the chip price, depending on sales volume. When Cisco acquired Radiata, it inherited the TLA and paid CSIRO approximately $900,000 in royalties over the next several years. Cisco stopped paying royalties under the TLA in 2007, when it discontinued the use of Radiata chips in its products. Cisco and CSIRO negotiated for several years regarding an ongoing license for the '069 patent, but could not reach agreement and CSIRO sued Cisco for infringement in 2011.
After a four-day bench trial, the District Court determined that the "reasonable" range for royalties for the '069 patent was between $0.90 (based on an "informal suggestion" made by Cisco's chief patent counsel during negotiations in 2005) and $1.90 (based on the maximum rate that CSIRO offered to potential licensees in 2003). CSIRO v. Cisco, 2014 WL 3805817 (E.D. Tex. 2014). (The Court made a slightly different calculation with respect to products sold by Cisco's Linksys subsidiary, but we will not consider that here). With these ranges in mind, the District Court developed a volume-based royalty table and assessed damages of approximately $16 million against Cisco. Cisco appealed, arguing that the District Court erred in three major regards: (1) by failing to begin its royalty analysis with the price of a Wi-Fi enabled chip, representing the smallest salable patent-practicing unit (SSPPU) in Cisco's Wi-Fi enabled products, (2) by failing to adjust its reasonable royalty analysis to account for the essentiality of the '069 patent to the 802.11 standard, and (3) by basing its royalty determination on the parties' negotiation positions rather than the TLA. The Federal Circuit found that the District Court erred as to points (2) and (3), vacating the decision below and remanding for recalculation of the damages award.
Apportionment and the Smallest Saleable Patent-Practicing Unit (SSPPU)
In CSIRO, the Federal Circuit reiterated the century-old rule, now embodied in Section 284 of the Patent Act, that patent infringement damages "must reflect the value attributable to the infringing features of the product, and no more." Ericsson, Inc. v. D-Link Sys., Inc., 773 F.3d 1201, 1226 (Fed. Cir. 2014). It went on to explain that "[t]his principle—apportionment—is the governing rule where multi-component products are involved" (slip op. at 10, internal quotations omitted). The Court noted, however, that the rule of apportionment may be supplemented by additional tools to help determine the incremental value of the patented invention. One of these tools is the SSPPU model: when "a damages model apportions from a royalty base, the model should use the smallest salable patent-practicing unit as the base" (slip op. at 12).
Cisco argued that the District Court erred by beginning its reasonable royalty analysis using rates derived from inter-party negotiations rather than the SSPPU. The Federal Circuit disagreed, holding that the SSPPU model was inapplicable in the case, as the District Court did not determine a royalty "base" (i.e., the price that is multiplied by a royalty expressed in percentage terms) at all. Instead, the Court used so-called per-unit royalties (i.e., a specific dollar amount per end product). As such, there was no call for use of the SSPPU model, and the District Court did not err by avoiding its use.
Comparable Licenses
As noted above, the District Court determined the applicable range of royalties on the '069 patent to be between $0.90 and $1.90 based on figures introduced by the parties at different stages during their licensing negotiations. Cisco argued, however, that the appropriate royalty range should be based on the rates set forth in the TLA entered into by Radiata and CSIRO, as to which Cisco later succeeded. Under the TLA, the royalty rates for Cisco products would have been $0.03 to $0.33 (rates for Linksys products would have been slightly different, but I will disregard these for the sake of simplicity).
The District Court rejected any application of the TLA in its reasonable royalty analysis, reasoning, among other things, that (a) the TLA was a related-party agreement between CSIRO and one of its former scientists, rendering it not comparable to the proposed arm's length agreement between Cisco and CSIRO, (b) the TLA was entered in 1998, long before any hypothetical negotiation between Cisco and CSIRO, and (c) the TLA royalty rates were based on the price of chips sold by Radiata rather than the value of the invention embodied by the '069 patent. To this last point, the District Court reasoned that "[b]asing a royalty solely on chip price is like valuing a copyrighted book based only on the costs of the binding, paper, and ink needed to actually produce the physical product. While such a calculation captures the cost of the physical product, it provides no indication of its actual value" (CSIRO, 2014 WL 3805817 at *11).
The Federal Circuit rejected most of the District Court's reasoning regarding the TLA, largely because Cisco and CSIRO renegotiated numerous terms of the TLA following Cisco's acquisition of Radiata. This renegotiation demonstrated both that the TLA did not embody a "special relationship" between CSIRO and the licensee (as Cisco presumably negotiated at arm's length) and the timing of the amendments coincided with any hypothetical negotiation that would have been conducted between Cisco and CSIRO. As for the District Court's discomfort with the TLA's use of chip prices as the base upon which royalties would be calculated, the Federal Circuit quoted its recent decision in Ericsson, 773 F.3d at 1228,
in which it held that a comparable license may not be excluded from the fact finder's consideration "solely because of its chosen royalty base." Given this reasoning, the Federal Circuit held that the District Court erred by excluding the TLA from its analysis and directed the Court on remand to "reevaluate the relevance of the as-amended TLA in its damages analysis" (slip op. at 22).
Interestingly, this case represents the second appellate decision this year in which the admissibility of comparable license agreements has been challenged in RAND royalty determinations. In the prior case, Microsoft v. Motorola, 795 F.3d 1024 (9th Cir. 2015), the Ninth Circuit was more deferential to the District Court's exclusion of potentially comparable license agreements. In Microsoft, the Circuit Court upheld the District Court's exclusion of three arm's length license agreements to which Motorola was a party for reasons including the fact that some agreements were entered into to settle or forestall litigation, they included patents other than the patents at issue, they included cross-licenses and they included royalty caps. It will be interesting to see how the Circuits reconcile their interpretations of this key evidentiary standard in future cases.
Impact of Standardization
Perhaps the most far-reaching implication of CSIRO arises from the Federal Circuit's holding regarding the impact of standardization on a patent. In the case, the District Court determined a "reasonable royalty" using the well-known framework established in Georgia-Pacific Corp. v. U.S. Plywood Corp., 318 F. Supp. 1116 (S.D.N.Y. 1970). Cisco argued that the District Court erred by failing to modify the Georgia-Pacific factors to account for the fact that the '069 patent was essential to the 802.11 standard. In particular, the Court failed to disregard any additional compensation that CSIRO might have been able to extract in a hypothetical negotiation solely as a result of the '069 patent's essentiality to the 802.11 standard. This additional compensation, Cisco argued, is not indicative of the incremental value of the patented technology, but of the significant costs that manufacturers would have to incur if forced to switch to an alternative technology (so-called "switching costs"). For this reason, such adjustments to the Georgia-Pacific factors were made by prior courts determining reasonable royalty rates for standards-essential patents (e.g., Microsoft, Ericsson, and In re Innovatio IP Ventures, LLC, 956 F.Supp.2d 925 (N.D.Ill. 2013)).
But CSIRO pointed to a significant distinction with these prior cases. As noted above, CSIRO agreed to license the '069 patent on RAND terms to manufacturers of 802.11a-compliant products. But by the time of CSIRO's suit, 802.11a was largely obsolete and represented only 0.03% of Cisco's accused products (slip op. at 8). Thus, CSIRO argued and the District Court agreed that CSIRO had no obligation to offer RAND terms to Cisco with respect to its products implementing later versions of 802.11. And because no RAND obligation was implicated, no adjustment to the Georgia-Pacific factors was warranted.
The Federal Circuit disagreed, holding that the incremental value of a standard-essential patent (SEP) should be determined independently of manufacturer switching costs, whether or not the SEP was RAND-encumbered (slip op. at 17). Citing Ericsson, the Court reasoned that "damages awards for SEPs must be premised on methodologies that attempt to capture the asserted patent's value resulting not from the value added by the standard's widespread adoption, but only from the technology's superiority" (id.) The Federal Circuit thus found that the District Court erred by failing to consider the extent to which the value of standardization may have impacted the calculated compensation range for the '069 patent, and remanded for further consideration of this issue.
Implications for RAND and Standards
The Federal Circuit's analysis of the third factor in CSIRO is sensible, but does raise some interesting questions about standards and SEPs. In rejecting CSIRO's argument that the royalty damages analysis should not be adjusted because the '069 patent was not RAND-encumbered, the Federal Circuit noted first that Ericsson distinguished between RAND-encumbered SEPs and SEPs generally (slip op. at 17). On this basis, the court reasoned that even though the '069 patent might not be encumbered by a RAND commitment, the court's reasonable royalty analysis must take into account the fact that the patent was a SEP (and thus correct for excess compensation that could be extracted based on broad industry adoption of the standard).
If this is the case, then what is the difference in the royalty payable with respect to a RAND-encumbered SEP and the royalty payable with respect to an unencumbered SEP? The result in CSIRO suggests that there is no difference at all. In the case of RAND-encumbered SEPs, the patent holder agrees to charge a "reasonable royalty", which the courts have calculated using a modified version of the Georgia-Pacific framework. But Section 284 of the Patent Act establishes a "reasonable royalty" as the baseline measure of damages for all patents. Accordingly, a similar reasonable royalty calculation, also using the Georgia-Pacific framework should be used for unencumbered SEPs. And, as held by the Federal Circuit in CSIRO, that calculation must avoid the inclusion of switching costs in the "reasonable" royalty.
In a recent paper, A Unified Framework for RAND and other Reasonable Royalties, 30 Berkeley Tech. L.J. 1447-1499 (2015), Richard Gilbert and I predict this result: namely, the convergence of reasonable royalty damages for RAND-encumbered and unencumbered patents. As we have written, and as the Federal Circuit has repeatedly confirmed, the appropriate measure of damages in patent cases, whether or not involving SEPs, is the incremental value of the patented invention to the product in which it is incorporated.
But if royalty rates for RAND-encumbered SEPs are no lower than royalty rates for unencumbered SEPs, then what is the point of making a RAND commitment? Does it have any effect at all? Professor Gilbert and I argue that RAND commitments are meaningful even without this royalty differential. Most importantly, a SEP holder that makes a RAND commitment severely limits its ability to obtain an injunction to prevent infringement by manufacturers of standardized products. Holders of unencumbered SEPs, on the other hand, have not committed to license their patents, and may not face the same hurdles to obtaining injunctive relief. Then, as predicted by Farrell, Lemley, Shapiro and others, they could use the leverage conferred by the threat of an injunction to extract a higher (unreasonable?) royalty from manufacturers of standardized products without having to resort to a judicial damages determination (which, as we have seen, will be limited to a "reasonable" royalty). This possibility has significant implications, particularly given the increasing acquisition and assertion of SEPs by patent assertion entities that do not make RAND commitments, a complex topic well beyond the scope of this note, but which I have written about here. For these and other reasons, RAND commitments, and the encouragement of RAND commitments by SSOs and market participants, will continue to play an important role in fostering standardization and innovation.
About Jason Rantanen
Jason is a Law Professor at the University of Iowa College of Law.	View all posts by Jason Rantanen →
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5 thoughts on "Guest Post by Prof. Contreras – CSIRO v. Cisco: The Convergence of RAND and non-RAND Royalties for Standards-Essential Patents"
Ned Heller says:
"[T]he appropriate measure of damages in patent cases, whether or not involving SEPs, is the incremental value of the patented invention to the product in which it is incorporated."
Damages? I think, rather, "reasonable royalty."
The appropriate measure is what the market will bear.
Value is indeed subjective and without the ability of the full measure of what a patent IS (a negative right to exclude), the same of abuse of "efficient infringement" must be zealously guarded against.
I do not know why the patent owner would have a standard essential patent without being part of the FRAND groups, but let us assume that one had a product on the market that was replaced by the product the subject matter of the FRAND obligation. I can see significant amounts of damage to the patent owner caused by the infringement of everybody in the FRAND group.
Ned Heller,
"I do not know why the patent owner would have a standard essential patent without being part of the FRAND groups."
That's possible because may be the patent owner did not take part in any process of the standard draft, in other words, other companies contributed a similar technology to the SSO without any notice that the technology had already been patented by some one else.
silicone valley girl says:
Isn't the call for FRAND licensing based upon the equitable notion that if you participate in having the standard that all must meet, then you should not be permitted to "gouge" those who meet the standard by rigorously enforcing your patent?
Assuming you agree with that premise, then if a patent is infringed by a product by complying with a standard and happens to be owned by somebody who had nothing to do with setting the standard free of those equitable restraints? Yes, that patent owner got lucky, but since when is getting lucky something to get criticized or penalized for?

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: How to pass mongodb in-built function as paramater in insert query using python I have a plan to create an index and delete it automatically. So according to the docs, it suggests ensureIndex function with createdAt parameter to do automatically. It works fine if I give the query directly in the mongo shell.
Now coming to the doubt, I have a list of entries with fields that are to be inserted in the specified collection using python. How should I give createdAt parameter which has a value new Date() which is a built-in in mongo so how to insert it?
In mongo shell:

db.test1.insert({ "createdAt": new Date(),    "logEvent": 2,    "logMessage": "Success!"})

Using python :
import pymongo
myclient = pymongo.MongoClient("mongodb://localhost:27017/")
mydb = myclient["sm"]
mycol = mydb["test1"]
d={}
d['logEvent']=2
d['logMessage']="Success!!"
d['createdAt']=new Date()
mycol.insert(d)

If i tried ,db.test1.find() displayed createAt as ISO date format
{ "_id" : ObjectId("5dca4d3a4845f6fcdd9bec84"), "createdAt" : ISODate("2019-11-12T06:12:10.942Z"), "logEvent" : 2, "logMessage" : "Success!" }

How to insert new Date()  using pymongo?

A: When you're in the Mongo Shell, you have a complete Javascript environment at your disposal, in which new Date is a valid expression. It is not in Python.
What you want is to pass a date type object with the current time from Python to Mongo. The pymongo driver understands how to translate a Python datetime object into a Mongo ISODate type. So just give it that:
from datetime import datetime

d['createdAt'] = datetime.now()

A: Mongodb uses the UTC timezone. So we need to pass UTC timezone formatted date object

    import pymongo
    from datetime import datetime
    myclient = pymongo.MongoClient("mongodb://localhost:27017/")
    mydb = myclient["sm"]
    mycol = mydb["test1"]
    d={}
    d['logEvent']=2
    d['logMessage']="Success!!"
    d['createdAt']=datetime.utcnow()
    mycol.insert(d)

Now it works and deletes the record automatically !!!

redpajama_set_name: RedPajamaStackExchange


lol...running and screaming...love it. Awesome, Aesome photos.
Scorpions are bad enough for me!
Sorry, but that's pretty funny! ;) Rats don't bother me. Either because I'm a scientist, and I've had to work with them, or because I live in Boston, so I see them with some regularity. Now, if I happened across a spider colony, I'd be paralyzed with fear!! It sounds like you have some great places to run.
Beautiful photos of apparently a pretty dirty place. :) Scary realization that you are out there alone sometimes, with no help coming.
Love the photo!! Ewww!! I hate rats! They're so dirty and gross!!
So funny! Something about being caught off guard makes it all so much scarier.
arm flapping attracts sharks as well - good that you got out of there!
Oh my- I would've flipped out!
I saw a relatively small rat one time and it completely freaked me out. Rats = gross.
Yikes! Mice I don't mind...rats are another issue! Great pics.
EWW! Glad your ok! WHERE THE HELL ARE THE NINGA TURTLES WHEN YOU NEED THEM!?
Haha, that must have been quite a sight...you two, not the rats. I have to dispose of a dead rat nearly every day as my cats catch them, bring them inside and leave them when they are dead. The photos are beautiful.
I didn't know I could run so fast!!! LMAO Our pace had to have been something like 8 minutes miles. Hopefully I'll see one at the marathon.
Running back to the car for your camera was so worth it... I love the pictures.
LOL! What? No rat pictures?


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text:
Lawsuit Claims Police Stripped Dead Crash Victim, Took Pictures of Her Naked Body
The family of a Chicago woman killed in a car crash has filed a lawsuit claiming police removed her clothing and took pictures of her naked body.
Jessica Mejia, 20, died on Dec. 31, 2009, when the vehicle she was riding in, driven by her boyfriend, spun out of control and hit a pole before rolling into a ditch.
The lawsuit claims officers from the Cook County Sheriff's office stripped Mejia's body in plain view of onlookers and took several photos of her on the side of the road, none of which were required by protocol, MyFoxChicago reported.
Department spokesman Steve Patterson denied the allegations and said despite being in regular contact with the family since the crash, the department "never before heard of the bizarre scenario this lawsuit describes."
"There were multiple independent witnesses on the scene that night and crime scene photographs taken which would have prevented the scenario described from happening. We're sorry for the family's loss," he said in a statement.
Mejia's boyfriend is awaiting trial on charges of reckless homicide and driving under the influence in the crash.
The lawsuit seeks more than $200,000 for "severe and extreme distress."
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Travel East on Rt. 120 to Greenbay Road. Turn left on Greenbay Road and travel north to Washington Street. Turn right on Washington to McAree.


text:
"Money talks, bull**** walks"
I note with interest – and concern – a step that's a very clear forecast of what lies ahead for consumers in the US economy.
Wells Fargo & Co said on Friday its second-quarter profit nearly halved as the bank set aside more funds to cover potential loan losses, while its mortgage lending business came under pressure from higher interest rates.
The fourth-largest U.S. bank reported profit of $3.1 billion, or 74 cents per share, compared with $6 billion, or $1.38 per share, a year earlier. Its total loan loss provisions were $580 million in the quarter, including a $235 million increase due to loan growth.
Under an accounting standard that took effect in 2020, banks must factor the economic outlook into loan loss reserves. Last year, the bank had released $1.6 billion from its reserves for loan losses as the economy rebounded from the pandemic.
Wells Fargo Chief Financial Officer Mike Santomassimo told reporters that retail and business customers remain strong, but the bank is prepared for a potential economic downturn.
"Things will probably get worse, but that's already included in the overall scenario analysis and the allowance level we have for the quarter," Santomassimo said.
Big bank executives have sounded cautious so far this earnings season, with JPMorgan Chase & Co's Chief Executive Jamie Dimon likening the macroeconomic environment to a coming "storm."
Wells Fargo isn't alone in increasing its reserves and provisions against default by its customers. Nevertheless, the numbers above are a bit mind-boggling. For a start, "its second-quarter profit nearly halved". Even with additional provisions against future losses, that's astonishing. Banks, like all other corporations, must pay dividends to their investors (shareholders). For Wells Fargo to say to its shareholders, "Sorry, your income from your investment in us has just been cut in half", and for them not to complain about it, speaks volumes. They're worried that if they don't make the investment in reserves right now, they'll be out a lot more money in future.
Secondly, this one bank has taken more than half a billion dollars of its own profits, in a single quarter, and allocated them as a reserve to write off debts – loans, mortgages, whatever – that it expects its customers to be unable to repay. What's it going to allocate next quarter? And the one after that? This is unlikely to be a one-off budgetary provision. It's the first major step in what's likely to become a longer-term major problem.
Finally, what does it say about Wells Fargo's expectations for the US economy? Clearly, the bank isn't greatly encouraged by what it sees "on the street". There's an old saying that "Money talks, bull**** walks". Well, one of the biggest banks in the USA is talking with its money, despite the glowing forecasts by our politicians that all will be well. It's treating those forecasts as something that needs to walk. That speaks volumes.
"He who has ears to hear, let him hear."
Published July 20, 2022 By Peter
Categorized as Uncategorized Tagged Business and Commerce, Current Events, Economics, Reality, Security
Let those with ears and eyes indeed.
Distractions is a carnivals stock in trade ALSO Politicians.
It's hard to pay attention to what is important when the Military Football game is on and "we're winning".
Mass shootings so far, ALL Known to the FBI and such, police standing down and allowing criminal behaviors.
Protect your family and trusted friends. Accept that betrayal is something that will happen given our Patriot Act "See something, Say something" and get rewarded for it behaviors.
The Famine wars is here, count any day of "Peace" a blessing to prepare for the storms on the horizon.
Pray for wisdom and act on it. Stop paying attention to the military football games friend.
As Arthor Sido's comment said so well, "What you do TODAY will determine who survives in 18 months".
A house built on sand……
This is yet another look at the grains slipping loose and sliding out from under our economy.
Barry Jones says:
The highlighted area indicsated the WF changed the Dividend for stock holders. It didn't so what are you referring to?
The vast majority of people I talk to have an amazing array of reasons (excuses) for not prepping. Ranging from I can't afford it to I don't want to think about it to you are just a pessimist. They don't want to face reality and they don't want to reallocate the money from new stuff and entertainment for themselves or their children.
I guess it is important to keep informing people as best I can, but like Peter on this blog, I'm preaching to the already converted or the intentionally deaf.
The ones that really drive me crazy are the ones that say they have land and a cabin many miles away in the roaring wilderness. They will chop wood and hunt to survive. Given their age and general health, they obviously have no idea how much wood will be needed. Shooting one deer a year isn't training for subsistence hunting. More importantly how are you going to get there. Anyone who has seen the evacuations for disasters like hurricanes has seen the utter chaos when everybody decides it is time to go and that is when the disaster is limited in scope. Will you be able to find clear roads and get sufficient fuel? How many opportunistic criminals will you have to get passed? Will you be able to deal with the groups that will see an isolated cabin as an opportunity?
These kind of questions get me labeled as annoying.
I think that the reachable have already been reached, the next level of the game will come when those who waken too late try to gather what little resources they can. The wise will have prepared and faded into the woodwork by then.
My part time gig to supplement my retirement is contacting delinquent mortgagees in person. In the past two months it has become nearly full time.
Exercise with less than half a back
A Polish-Ukrainian "commonwealth"? Now there's a thought…


RESIDENTS OF the State, especially the Capital are being subjected to vagaries of nature with the weather blowing hot and cold during the last few days.
Although December has begun, the State has not yet experienced typical winter conditions. Just a day after day temperatures in most parts of State slumped considerably under the influence of clouding and sporadic rains, the mercury suddenly shot up on Monday all across the State.
High day and night temperatures were recorded with Bhopal recording a particularly high minimum temperature of 17.9 degrees Celsius on Sunday night - almost six degrees above normal of the season. Minimum temperatures between 10 and 12 degrees Celsius are recorded in the Capital during this time of the year.
degrees above Saturday's maximum of 22.5 degrees.
In other parts of the State too, Indore recorded maximum temperature of 30.8 and minimum of 14.5, Gwalior – 30.1 and 13.2 and Jabalpur – 30.3 and 15.0. The lowest minimum was recorded at Khajuraho at 12.0 degrees. This is quite in contrast with last week's scenario when the minimum had plunged down to 8 degrees in various parts of the State.
Regional Meteorological Centre Director Dr D P Dubey said this scenario was being experienced because of the change in wind pattern under the influence of a weather system prevailing over Rajasthan.
This weather system has caused the wind direction over entire Madhya Pradesh to change to southerly, thus pulling in warm air. The same weather pattern is causing rains in Jammu and Kashmir and Rajasthan but leading to high temperatures in Madhya Pradesh.
Citizens can keep their heavy woollens stashed away as similar weather is expected to prevail in the State for at least another two days.
The weather is expected to change only when clouding is completely cleared in Western and Northern India. The Capital has been witnessing real topsy-turvy weather conditions, particularly in day temperatures over the past few days.


text:
Restrict my search to Division of Research
Division of Research
National Centers
Consortiums
Research and Scholarship Committee
UH Research Deans
Notable Faculty
Limited Submissions
Internal Awards
Texas Research Incentive Program (TRIP)
Center Requests For Proposals (RFPs)
Internal Funding Terms & Conditions
Post-Award Guidance
Research Contract Office
PeopleSoft Grants
PI Portal
Uniform Guidance A-81
Responsibilities Matrix
Procedures, Policies and Guidelines
Negotiated Indirect Rates
ICON Online System
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Conflict of Interest Certification in Research
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DOR Forms
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R&I Forum
Email the DOR
Send Kudos
Travel/Fly America Act
Transmittals
Beverly Rymer
Executive Director, Office of Contracts and Grants
brymer@uh.edu
FAQs: Travel/Fly America Act
Fly America Act Waiver Checklist (DOC)
Can airfare be booked using a travel website or a travel agent?
The best way to book the flight is to go directly through a U.S. flag air carrier. If you have to use a travel website or travel agency, verify that the flights you are purchasing are being issued by the U.S. flag air carrier. University-contracted travel agencies may not be aware of the source of funds you are using to pay for the travel and the need to fly on a U.S. flag air carrier. You should let the agency know in advance.
Who is required to use a U.S. flag air carrier?
All individuals who will be paid or reimbursed for travel costs from a federally funded project cost center.
What is a Code Share flight?
A code share flight involves an agreement in which a ticket is issued by a U.S. flag air carrier, although the seat is on a plane operated by a non-U.S. flag air carrier. For example, United Airlines may sell a ticket that includes some legs of a trip to Germany on a Lufthansa plane. To be compliant for federal funds, the ticket or itinerary must show the U. S. flag air carrier's airline code next to the flight number for each leg of the trip. If the flight number had an airline code of UA for United Airlines, the ticket would be compliant, even if one or more legs of the trip were "Operated by Lufthansa." If the ticket had an airline code of LH for Lufthansa, the ticket would not be compliant and could not be charged to federal funds.
Where can I find the airline designator codes?
These are a couple websites to go to for determining the country of origin for air carriers:
http://www.airlinecodes.co.uk/airlcodesearch.asp
http://www.tvlon.com/resources/airlinecodes.htm
May I travel by a foreign flag air carrier if the cost of my ticket is less than traveling by a U.S. flag air carrier?
No. The Fly America Act specifically states that cost cannot be considered as a factor.
Scenario Question
Flight tickets within the U.S. and on U.S. air carriers were issued by a U.S. air carrier, and flights to and/or within foreign countries were on foreign carriers with tickets issued by a U.S. carrier. Can the entire trip be charged to the federal funds?
No. Even when the traveler is in a foreign country and is traveling from one foreign location to another, U.S. air carriers would still need to be utilized, if available. A waiver will be needed to document why service by U.S. flag air carrier(s) is not available, or why it was necessary to use a foreign air carrier.
What is bilateral or multilateral air transportation agreement?
This is the same as the Open Skies Agreements, which allows federally funded airline tickets to be purchased on foreign air carriers that belong to countries that have an agreement with the U.S. However, the following conditions must exist:
A General Service Administration (GSA) city-pair contract does not exist
Travel is not funded by the Department of Defense
The foreign carrier is a member airline of the European Union, Australia, or Switzerland.
What is a (GSA) City Pair agreement?
Each year, under the Airline City Pair Program, the GSA Federal Supply Service awards contracts for air transportation for travelers on official government travel. The contracts are competitively awarded based on the best overall value to the government. Government employees are required to use the U.S. flag airlines if a city-pair agreement exists, without regard as to whether a bilateral or multilateral air transposition agreement exist with that country.
What is considered an involuntarily rerouting of a traveler on a foreign air carrier?
An involuntary rerouting means that the initial airline ticket or itinerary was booked with no foreign carrier listed. However, while traveling you were asked to change carrier due to unforeseen circumstances.
Sexual Misconduct Reporting and Awareness


President Donald Trump on Tuesday morning said the U.S. would be at war with North Korea if it weren't for him.
But the continued talks come as reports from USA intelligence agencies reveal that Pyongyang is continuing to work on its nuclear and intercontinental ballistic missile systems while engaging in negotiations.
Sen. Lindsey Graham (R-S.C.), a member of the Armed Services Committee, told NBC on Sunday that if the intelligence reports are true, North Korea is "saying one thing and doing another, nobody should be surprised".
White House spokeswoman Sarah Sanders told a news briefing that the United States was "continuing to make progress" in talks with North Korea.
The Washington Post reported yesterday that North Korea was considering ways to hide its nuclear weapons and secret production facilities.
Speaking on CBS' "Face the Nation" on Sunday, White House national security adviser John Bolton said that Pompeo would be "discussing this with the North Koreans in the near future about really how to dismantle all of their WMD and ballistic missile programs in a year".
Recent revelations include an NBC News report that said U.S. intelligence agencies believe the North has increased its production of enriched uranium fuel for nuclear weapons at multiple secret sites in recent months.
After the meeting, Trump tweeted that there was no longer a nuclear threat from the country.
Bolton said Pompeo will present Kim with a USA plan for complete Korean denuclearization within a year during his upcoming trip to North Korea. "There is absolutely unequivocal evidence that they are trying to deceive the U.S". The president tweeted last month that there "was no longer a nuclear threat" posed by North Korea.
Separately, he told "Fox News Sunday" that "we are using the full range of our capabilities to understand what North Korea is doing".
USA officials, led by Ambassador Sung Kim, met with their North Korean counterparts Sunday at the demilitarized zone that divides North and South Korea, the first such meeting since the Singapore summit.
North Korea appears to be finalizing the expansion of a key ballistic missile manufacturing site, according to an analysis of satellite imagery by researchers in the United States. "Kim may want to give up his weapons but the generals may not". According to some reports, Pompeo may make another visit to North Korea this week for more talks.
Experts say Pompeo will have to coax significant reciprocal steps from North Korea that would firmly lock the country into a process of disarmament to justify the suspension of the U.S.


EVERYTHING COMES TO HEAD! G.I. JOE's high seas rescue mission has gone from bad to worse. Now, each team present must regroup and rally in hopes of stopping the BLACK MAJOR and his RED SHADOWS once and for all. The action heats up in G.I. JOE: A REAL AMERICAN HERO #190!


I started my paid culinary life creating recipes for a steam oven. So imagine my delight to find out there was a steaming device that came with the Thermomix. I quickly converted a lot of my existing recipes to the Varoma. I was in love. Cheesecake with even texture, soups with steamed veggies over the top. Stuffed chicken rolls steamed to tender juicy perfection, pork belly, yes, fragrant pork belly steamed before hitting the high heat of a grill to get the crackling organized. Roast lamb…not kidding. All of these amazing dishes with fantastic results using one of the most ancient cooking methods we know of.
Apparently steaming is a gentler kind of cooking, and actually will preserve flavour, nutrients and of course moisture in your foods, as well as being very conservative on energy.
Don't underestimate using the Varoma to create amazing steamed meat dishes. I love a main meal salad that utilises steamed chicken or pork. I suggest marinating the meat of choice for a few hours prior to steaming. Try and have the meat in equal sized portions prior to steaming, and line the tray or dish with wet baking paper as this saves on clean up later. I also have a fantastic bolognaise sauce on my website that steams the minced beef over the sauce, thus preserving texture and increasing the volume capacity. You add the fully cooked beef to the sauce when ready to serve. Meatballs will all benefit from being partially steamed, either over the sauce they are going to be served with, or over water, prior to being finished off in a hot oven. The steaming will help them keep their shape…so think about it…rissoles, skinless sausages. All easier thanks to the Varoma.
As the self proclaimed Queen of Desserts, I love using the Varoma for so many of my dessert recipes. Many versions of my steamed cheesecake are out there. The texture is smooth, creamy, no cracks through the middle and so easy to recreate those perfect results.
And do I need to mention Sticky Date Pudding? Surely not. Or Christmas pudding? Or self saucing puddings? I'm stopping now.
I mention noodles because the most traditional method for making Fresh Rice Noodles is steaming over a very large wok on an open flame. No need to go to that trouble. Use your Varoma. The very simplest method ever.
It almost goes without saying…but I'm going to say it anyway. The Varoma is ideal for veggie steaming. Of all kinds.
If you are keen to make a potato bake, it is always wise to pre-steam your potatoes in the Varoma first. I steam my potatoes for mash in the Varoma prior to letting them dry a little, then mashing them in the TM bowl with all the usual suspects; cream, butter, cheese, herbs etc.
Think beyond just the usual. You can pre-steam an eggplant, or pumpkin, or sweet potato, stuff them, and finish them off in the oven to get that caramelized taste we all love. Think outside the steam!
OK. This is Varoma Therapy at it's finest. Fish is so wonderful when perfectly steamed. I simply wrap fish fillets or mixed seafood in foil or baking paper that has had flavoured butters, herbs, lemon slices, seasoning added and steam away. You can even layer up with the fish parcels and cook rice in the simmering basket as you steam the fish. Add a salad and a glass of white. Dinner is served. In probably around 15-20 minutes. That easy.
Last but not least, the results you can get from partially steaming a baguette before transferring it to the hottest oven possible are amazing. Chewy golden glossy crust, all because of a bit of steam.
Then there are all the Asian dim sum style 'bread 'options you can create. Think Pork Buns, dumplings of all kinds.
And lest I leave you hanging, of course I have recently brought out my Varoma Therapy cookbook, available now as either an ebook (Free to Insiders) or a print book. GO GET IT and become your own Varoma Therapist.
So get your Varoma out of the cupboard, dust it off and start steaming your kitchen up. It is a versatile addition to your cooking arsenal, as well as offering the perfect way to increase the yield to feed the hungry hoardes that no doubt live at your place as well as mine. Varoma Therapy. We all need it.
How Chocolate Makes you Calmer, and More Creative!
Should I or Shouldn't I upgrade to the new Thermomix 6 model?
Green Smoothie School. No Recipe Required!
My Top Seven meal hacks for busy families to save you time and money!
8 Christmas Dishes You Can't Do Without.
Drop Smart Kitchen Scale and ME! A new Partnership!
Grazing Tables; The How To Article.
My Top Ten Favourite Ingredients I always have in my available in my pantry.
Three Smoothies That are Better Than Dessert!
Meringue Tips; Or The Secret to Fluff!
Matcha Madness; 5 Reasons Why Everyone is going Mad for Matcha!
NEW ebook: Top 25 Ice Cream Recipes and more!
My new Christmix mini ebook.
Merry Christmix One and Merry Christmix Too...the printed books!
Ebook: Merry Christmix Too, the new and biggest thermomix ebook from ME to YOU!
Thermomix Ebook: What's for Breakfast?
The New Ebook - What's For Breakfast?


HomeHomeFPI has achieved second place for Innovation Driven R&D Outsourcing!
We are so proud to announce that FP Instruments has received an award for Innovation Driven Companies, which is given to the residents of Technology Parks located in Poland.FP Instruments was recognized by the Association of Business and Innovation Centres in Poland as an Innovation Driven R&D Company. FPI achieved second place in InnoFirma – Parkowe Orły 2015 contest, mainly for its significant growth potential, most innovative product and ability to operate on international markets.We have competed against more than 30 companies that deliver innovative products and services. This will certainly nurture our passion for electronics development as we are a team of developers with research background. This competition was organised by the Polish Association for Innovation and Entrepreneurship for the 5th year in a row, and it was part of the XXVI National Conference held in Poznań.


Title says is all. That's 21:00CET, 19:00GMT, 15:00EST, 14:00CST, 13:00MST, 12:00PST.
A small twist of lemon?
Break's over. I fixed the oob. You're back in business.
Excellent news in both cases.
Mm, yes, SoW OOB files do not like double quotes, so I can't have a RA 5.5" howitzer, I have to have 5.5in howitzers.
The Baden Fusilier error I do not understand. I think this sprite set was possibly not implemented correctly in Gunships mod. I notice that it has never been used in any OOB. I was going to use it to add another uniform variant but I can easily do without it.
Kevin, all those other errors involving sounds and graphics your installation threw up were due to Reb Bugler's toolbar mod which I have incorporated into my main Nap Mod... so its now part of the Peninsular Mod.
I think almost everyone uses RBs toolbar now... except you!
The trouble with RBs toolbar is it has to be incorporated into the main game if you want to host. A mod design fault I know but we're stuck with it; you are just going to have to use it.
I very much doubt we can fight the battle tonight anyway as no French players have signed up other than Mark, unfortunately.
We will definitely do it next weekend. That gives Mark a week to drum up some support!
That's a shame - I'll try to notify my team when I get home so not TOO many of us show up at midnight expecting mayhem.
At this stage I don't know my availability next weekend, so it may make sense to do one of the other battles first.
I plan also to make Saturday's game. As Kevin suggests, it should serve well as a dry run for historical map compatibility.
Sounds as though morale is waning among the croissant-soldiers -- and so soon after visiting our lovely Iberia! Perhaps they are unaware of the proper welcome soon to be administered?
I find Reb's toolbar an abomination. Does anyone really use all those buttons? I suggest changing all the sounds to stock ones, or moving them to a sound folder in your mod. That way players can decide which toolbar to use.
We definitely need a discussion about toolbars, and yes I use many of the buttons. I find them quite useful. Many I don't but no other toolbar has the buttons I *do* use. I generally play the game from the toolbar, I hardly ever send orders via couriers to my AI subordinates or peers.
I like the rocker-switch effect with 2 functions built into one button, its a very cool feature and packs far more functions into a small space. Its something I could do without if enough space could be found for all the buttons without that effect though.
We should set up a toolbar working party and design a Nap one that has the buttons we all need.
The sounds are not stock because they come from Gunships mod. Is my guess. Not sure.
Could we all meet earlier than 20:00 UK time on Saturday to do various installation tests on some vanilla maps?
I will miss todays game as we are playing boardgames. Hafe fun!


Basava International School | Dwarka New Delhi : Top School in Dwarka.
Nursery admission 2019-20 Draw of lottery will be held on 31 January 2019 at 3 PM in school permises.
Set up by the KLE Society of Karnataka, the school recognizes the need for quality education for the over all growth of children to face a competitive world. KLE Society runs about 290 educational institutions including well known-medical and engineering colleges around the country. This is one of the largest networks of institutions for higher studies and technical education in India. KLE will complete 100 years of its service in 2014.
At Basava International School we have a clear idea about our role in the children's journey into the future. We want our students to boldly meet the challenges of a brave new world imbibing the values of a 5000-year-old civilization. That has been the Indian experience so far. We do not want our students to be merely successful but tell them to make an impact.
Life skill programmes have been integrated into the school curriculum. Children at Basava learn to take life in their stride. The school promotes workshops to ensure that each child is bold and is embeded or empowered with fantastic success traits.
School offers a wide range of educational & sports facilities.
Principal : 9 :00 am to 11 :00 am, Teachers : Anytime within school hours. GM : Anytime in school hours.
Our students study and develop in a pollution & stress free environment.


text:
package org.apache.lucene.search.spell;

import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.io.Reader;

import org.apache.lucene.util.BytesRef;
import org.apache.lucene.util.BytesRefIterator;
import org.apache.lucene.util.IOUtils;

/**
 * Dictionary represented by a text file.
 *
 *
Format allowed: 1 word per line:

 * word1

 * word2

 * word3

 */
public class PlainTextDictionary implements Dictionary {

  private BufferedReader in;

  /**
   * Creates a dictionary based on a File.
   *

   * NOTE: content is treated as UTF-8
   */
  public PlainTextDictionary(File file) throws IOException {
    in = new BufferedReader(IOUtils.getDecodingReader(file, IOUtils.CHARSET_UTF_8));
  }

  /**
   * Creates a dictionary based on an inputstream.
   *

   * NOTE: content is treated as UTF-8
   */
  public PlainTextDictionary(InputStream dictFile) {
    in = new BufferedReader(IOUtils.getDecodingReader(dictFile, IOUtils.CHARSET_UTF_8));
  }

  /**
   * Creates a dictionary based on a reader.
   */
  public PlainTextDictionary(Reader reader) {
    in = new BufferedReader(reader);
  }

  @Override
  public BytesRefIterator getWordsIterator() throws IOException {
    return new FileIterator();
  }

  final class FileIterator implements BytesRefIterator {
    private boolean done = false;
    private final BytesRef spare = new BytesRef();
    @Override
    public BytesRef next() throws IOException {
      if (done) {
        return null;
      }
      boolean success = false;
      BytesRef result;
      try {
        String line;
        if ((line = in.readLine()) != null) {
          spare.copyChars(line);
          result = spare;
        } else {
          done = true;
          IOUtils.close(in);
          result = null;
        }
        success = true;
      } finally {
        if (!success) {
          IOUtils.closeWhileHandlingException(in);
        }
      }
      return result;
    }
  }
}

redpajama_set_name: RedPajamaGithub


text:
\section{Introduction}
A {\em spanning tree} of a connected graph $G$ is a subgraph that
includes all the vertices of $G$ and is a tree.  Spanning tree is a
fundamental concept in graph theory, and finding a spanning tree of
the input graph is a routine step of graph algorithms, though it
usually induces no extra cost: most algorithms will start from
exploring the input graph anyway, and both breadth- and
depth-first-search procedures produce a spanning tree as a byproduct.
However, a graph can have an exponential number of spanning trees, of
which some might suit a specific application better than others.  We
are hence asked to find constrained spanning trees, i.e., spanning
trees minimizing or maximizing certain objective functions.  The most
classic example is the minimum-weight spanning tree problem (in
weighted graphs), which has an equivalent but less known formulation,
i.e., maximum-weight spanning tree.  Other constraints that have
received wide attention include minimum diameter spanning tree
\cite{hassin-95-minimum-diameter-spanning-tree}, degree constrained
spanning tree~\cite{konemann-02-degree-bounded-mst,
  konemann-05-degree-bounded-mst}, maximum leaf spanning tree
\cite{lu-98-approximate-maximum-leaf-spanning-tree}, and maximum
internal spanning tree \cite{salamon-10-thesis-mist}.  Unlike the
minimum-weight spanning tree problem
\cite{karger-95-randomized-linear-mst}, most of these constrained
versions are NP-hard \cite{ozeki-11-survey-spanning-trees}.

The optimization objective we consider here is to maximize the number
of internal vertices (i.e., non-leaf vertices) of the spanning tree,
or equivalently, to minimize the number of its leaves.  More formally,
the {\em maximum internal spanning tree} problem asks whether a given
graph $G$ has a spanning tree with at least $k$ internal vertices.
Containing the Hamiltonian path problem as a special case, it is
clearly NP-hard.  This paper approaches it by studying {kernelization
  algorithms} for its parameterized version; here the parameter is
$k$, and hence we use the name {\em $k$-\ist}.  Given an instance $(G,
k)$ of $k$-\ist, a {\em kernelization algorithm} produces in
polynomial time an ``equivalent'' instance $(G', k')$ such that $k'
\leq k$ and that the {\em kernel size} (i.e., the number of vertices
in $G'$) is upper bounded by some function of $k'$.  Prieto and Sloper
\cite {prieto-03-mist} presented an $O(k^3)$-vertex kernel for the
problem, and improved it to $O(k^2)$ in the journal version \cite
{prieto-05-mist}.  Fomin et al.~\cite{fomin-13-kernel-mist} crafted a
very ingenious reduction rule, and showed that a simple application of
this rule is sufficient to yield a $3k$-vertex kernel.  Answering a
question asked by Fomin et al.~\cite{fomin-13-kernel-mist}, we further
improve the kernel size to $2k$.
\begin{theorem}\label{thm:main-2}
  The $k$-\ist\ problem has a $2k$-vertex kernel.
\end{theorem}

We obtain this improved result by revisiting the reduction rule
proposed by Fomin et al.~\cite{fomin-13-kernel-mist}.  A nonempty
independent set $X$ (i.e., a subset of vertices that are pairwise
nonadjacent in $G$) as well as its neighborhood are called a {\em
  reducible structure} if $|X|$ is at least twice as the cardinality
of its neighborhood.  To apply the reduction rule one needs a
reducible structure.  Indeed, we are proving a stronger statement that
implies Theorem~\ref{thm:main-2} as a corollary.
\begin{theorem}\label{thm:main}
  Given an $n$-vertex graph $G$, we can find in polynomial time either
  a spanning tree of $G$ with at least $n /2$ internal vertices, or a
  reducible structure.
\end{theorem}

The observation in \cite{fomin-13-kernel-mist} is that the leaves of a
depth-first-search tree $T$ are independent.  Therefore, if the graph
has more than $3 k - 3$ vertices, then either the problem has been
solved (when $T$ has $k$ or more internal vertices), or the set of (at
least $2k - 2$) leaves of $T$ will be the required independent set.
It is, however, very nontrivial to find a reducible structure when $2k
< n < 3 k - 3$, and this will be the focus of this paper.  We first
preprocess the tree $T$ using a greedy procedure that applies a
sequence of local-exchange operations to increase the number of its
internal vertices.  After a local optimal spanning tree is obtained,
we show that if it has more leaves than internal vertices, then a
subset of its leaves and its neighborhood make the reducible
structure.  We apply the reduction rule of \cite{fomin-13-kernel-mist}
to reduce it and then repeat the process, which terminates on either a
$2k$-vertex kernel or a solution.

It is interesting to point out that our kernelization algorithm will
never ends with a NO situation, which is common in kernelization
algorithms in literature.  It either returns a trivial YES instance,
or continuously reduces the graph until it has a spanning tree with at
least half internal vertices.  This also means that our kernelization
algorithm does not rely on the parameter $k$.  One should be noted
that to further improve a $2k$-vertex kernel for a graph problem seems
to be a very challenging, if possible, undertaking: more and more such
kernels have appeared in literature, which have stubbornly withstood
all subsequent attacks, however hard they were.

Priesto and Sloper~\cite{prieto-05-mist} also initiated the study of
parameterized algorithms (i.e., algorithms running in time
$O(f(k)\cdot n^{O(1)})$ for some function $f$ independent of
$n$)\footnote{Following convention, we use the $O^*(f(k))$ notation to
  suppress the polynomial factor $n^{O(1)}$ in the running time.}  for
$k$-\ist, which have undergone a sequence of improvement.  Closely
related here is to the {\em $k$-\iob} problem, which, given a {\em
  directed} graph $G$ and a parameter $k$, asks if $G$ has an
out-branching (i.e., a spanning tree having exactly one vertex of
in-degree $0$) with at least $k$ vertices of positive out-degrees.  It
is known that any $O^*(f(k))$-time algorithm for $k$-\iob\ can solve
$k$-\ist\ in the same time, but not necessarily the other way.  After
a successive sequence of studies \cite{gutin-09-iob, cohen-10-iob,
  fomin-12-sharp-separation, shachnai-14-representative-families,
  daligault-11-dissertation}, the current best deterministic and
randomized parameterized algorithms for $k$-\iob\ run in time
$O^*(6.86^k)$ and $O^*(4^k)$ respectively, which are also the best
known for $k$-\ist.  Table 1 summarizes the history of this line of
research.

\begin{table}[h]
  \caption{Known parameterized algorithms for problems $k$-\iob\ and  $k$-\ist\
    (note that an algorithm for the former  applies to the later as well).}
  \medskip
  \centering
\begin{tabular}{l |l r r}
  \toprule
  Problem & Running time & Reference & Remark
  \\
  \midrule

  & $O^*(k^{O(k)})$ & Gutin et al.~\cite{gutin-09-iob} &
  \\
  $k$-internal  & $O^*(55.8^k)$ & Cohen et al.~\cite{cohen-10-iob} &
  \\
  out-branching & $O^*(16^{k+o(k)})$ & Fomin et
  al.~\cite{fomin-12-sharp-separation} &
  \\
  & $O^*(4^k)$ & Daligault and Kim \cite{daligault-11-dissertation} &
  \emph{randomized}
  \\
  & $O^*(6.86^k)$ & Shachnai and
  Zehavi~\cite{shachnai-14-representative-families} &
  \\
  \midrule

  & $O^*(k^{2.5 k})$ & Priesto and
  Sloper~\cite{prieto-05-mist} &
  \\
  $k$-internal & $O^*(2.14^k)$ & Binkele-Raible et
  al.~\cite{binkele-raible-13-mist} &   \emph{cubic graphs}
  \\
  spanning tree  & $O^*(8^k)$ & Fomin et al.~\cite{fomin-13-kernel-mist} &
  \\
  & $O^*(4^k)$ & {This paper} &
  \\
  \bottomrule
\end{tabular}
\end{table}

The $O^*(4^k)$-time {\em randomized} algorithm for $k$-\iob\
\cite[Theorem 180]{daligault-11-dissertation} was obtained using a
famous algebraic technique developed by Koutis and Williams
\cite{koutis-09-group-algebras-for-fpt}, which, however, is very
unlikely to be derandomized.  As a corollary of
Theorem~\ref{thm:main-2}, we obtain an $O^*(4^k)$-time deterministic
algorithm for $k$-\ist,---it suffices to apply the $O^*(2^n)$-time
algorithm of Nederlof \cite{nederlof-13-polynomial-space-by-PIE} to
the $2k$-vertex kernel produced by
Theorem~\ref{thm:main-2},---matching the running time of the best
randomized algorithm

\begin{theorem}\label{thm:alg}
  The $k$-\ist\ problem can be solved in time $O^*(4^k)$.
\end{theorem}
It remains an open problem to develop a deterministic $O^*(4^k)$-time
algorithm for $k$-\iob.  Note that the minimum spanning tree problem
has been long known to be solvable in randomized linear time
\cite{karger-95-randomized-linear-mst}, while a deterministic
linear-time algorithm is still elusive.  As a final remark, there is
also a line of research devoted to developing approximation algorithms
for maximum \ist\ \cite{prieto-03-mist, knauer-09-approximate-mist,
  salamon-09-approximate-mist, salamon-08-mist}.  In a companion paper
\cite{li-14-approximate-mist}, we have used a similar local-exchange
procedure to improve the approximation ratio to $1.5$.

\section{A greedy local search procedure}

All graphs discussed in this paper shall always be undirected and
simple, and the input graph is assumed to be connected.  The vertex
set and edge set of a graph $G$ are denoted by $V(G)$ and $E(G)$
respectively.  For a vertex $v\in V(G)$, let $N_G(v)$ denote the
neighborhood of $v$ in $G$, and let $d_G(v) := |N_G(v)|$ be its degree
in $G$.  The neighborhood of a subset $U\subseteq V(G)$ of vertices is
defined to be $N_G(U) = \bigcup _{v \in U} N_G(v) \backslash U$.  A
tree $T$ is a {\em spanning tree} of a graph $G$ if $V(T) = V(G)$ and
$E(T)\subseteq E(G)$; edges not in $T$, i.e., $E(G)\setminus E(T)$,
are {\em cotree edges} of $T$.  A vertex $u\in V(T)$ is a {\em leaf}
of $T$ if $d_T(u)=1$, and an {\em internal vertex} of $T$ otherwise;
let $L(T)$ and $I(T)$ denoted the set of leaves and the set of
internal vertices of $T$ respectively.  An internal vertex $u$ of $T$
is a {\em branchpoint} if $d_T(u)\geq 3$.  Let $I_3(T)$ denote the set
of branchpoints of $T$, and let $I_2(T)$ denote other internal
vertices (having degree $2$ in $T$); the three vertex sets $L(T),
I_2(T),$ and $I_3(T)$ partition $V(T)$.

Since $|I(T)| = |V(T)| - |L(T)|$, to maximize it is equivalent to
minimizing the number of leaves.  Also connecting leaves and internal
vertices, especially branchpoints, of a tree $T$ is the following
elementary fact:
\[
|L(T)| - 2 = \sum_{v\in I(T)} \left( d_T(v) - 2 \right) = \sum_{v\in
  I_3(T)} \left( d_T(v) - 2 \right).
\]
Therefore, informally speaking, we need to decrease the number and
degrees of branchpoints.  We start from an arbitrary spanning tree $T$
of $G$.  We may assume that $T$ is not a path, (as otherwise the
problem has been solved,) to which we apply some local exchanges to
increase the number of internal vertices of $T$.  By a local exchange
we mean replacing an edge in $E(T)$ by a cotree edge of $T$ .  Recall
that for any pair of vertices $u,v$ in a tree $T$, there is a unique
path from $u$ to $v$, denoted by $P_T(u,v)$; if $u v$ is a cotree edge
of $T$, then the length of $P_T(u,v)$ is at least two.  To maintain a
tree, a cotree edge $u v$ can only replace a tree edge in $P_T(u, v)$.
Since our purpose is to eliminate leaves, we will be only concerned
with cotree edges incident to leaves of $T$.  The first exchange rule
is self-explanatory; here the existence of the branchpoint is ensured
by the assumption that $T$ is not a path.

\begin{exrule}[\cite{prieto-03-mist}]\label{rule:1}
  If there is a cotree edge connecting two leaves $l_1$ and $l_2$ of
  $T$, then find an edge $u v$ from $P_T(l_1,l_2)$ such that $u$ is a
  branchpoint, and substitute $l_1 l_2$ for $u v$ in $T$.
\end{exrule}
After an exhaustive application of Rule~\ref{rule:1}, all leaves in
the resulting tree are pairwise nonadjacent.  Henceforth we may assume
that each cotree edge is incident to at least one internal vertex.  We
remark that Fomin et al.~\cite{fomin-13-kernel-mist} achieved this by
using depth-first-search tree at first place.  We can surely use the
same way to get the initial spanning tree, but we still need
Rule~\ref{rule:1}, as later operations of the other exchange rule to
follow may introduce cotree edges connecting leaves that are
originally not.

\begin{definition}
  A cotree edge of $T$ is {\em good} if it connects a leaf $l$ and an
  internal vertex $w$ of $T$.  We say that $l w$ {\em crosses} every
  edge in the path $P_T(l, w)$.
\end{definition}
For notational convenience, when referring to a good cotree edge $l
w$, we always put the leaf $l$ first, and when referring to an edge $u
v$ crossed by it, we always put the vertex closer to $l$ first; hence,
$P_T(l,w)$ can be written as $l \cdots u v \cdots w$.  We would like
to point out that the same edge $u v$ can be crossed by two different
cotree edges and they may be referred to by different orders.

Let us consider the impact of a substitution on the involved vertices.
We will avoid the tree edge incident to $l$ but we do allow $v = w$,
and hence there are either three of four vertices involved.  The two
vertices of the cotree edge are clear: $l$ always becomes an internal
vertex, and $w$ always remains internal (independent of $w = v$ or
not).  On the other hand, $u$ and $v$ will remain internal if they are
in $I_3(T)$, but one or both of them may become leaves if they are in
$I_2(T)$ (with the only exception $v = w$).  Although some operation
does not increase, or even decreases, the number of internal vertices,
by switching vertices in $L(T)$ and $I(T)$, it may introduce cotree
edge(s) between leaves in the new tree, which enable us to
subsequently apply Rule~\ref{rule:1} and serve our purpose.  This is
never the case for the first edge in $P_T(u, v)$ and hence we avoid
it.  This observation is formalized in the next reduction rule, for
which we need a technical definition that characterize those vertices
in $I_2(T)$ that can participate in the aforementioned successive
exchanges.

\begin{definition}
  All vertices in $I_3(T)$ are {\em detachable}.  A vertex $w\in
  I_2(T)$ is {\em detachable} if there exists a good cotree edge $l w$
  of $T$ satisfying at least one of the following:
  \begin{enumerate}[(1)]
  \item $P_T(l,w)$ visits a branchpoint, or
  \item some vertex $v$ in $P_T(l,w)$ is incident a good cotree edge
    $l'v$ of $T$ with  $l' \ne l$.
  \end{enumerate}
\end{definition}
Let $D(T)$ denote the set of detachable vertices of $T$, and let
$D_2(T)$ denote those detachable vertices in $I_2(T)$.  Then $ D(T) =
D_2(T) \cup I_3(T) \subseteq I(T)$.  Note that the vertex $v$ in
item (2) of the definition of $D_2(T)$ necessarily has degree two, and
possibly $v = w$.

\begin{exrule}\label{rule:2}
  Let $u v$ be an edge crossed by a good cotree edge $l w$ of $T$.
  Substitute $l w$ for $u v$ if any of the following is true.
  \begin{itemize
  \item $u\in I_3(T)$, and
    \begin{enumerate}[(a)]
    \item\label{case2} $v = w$ or $v\in I_3(T)$; or
    \item\label{case3} $v\in D_2(T)$ and there is a good cotree edge
      $l_v v$ with $l_v \ne l$.
    \end{enumerate}
  \item $u\in D_2(T)$, there is a good cotree edge $l_u u$ with
    $l_u \ne l$, and
    \begin{enumerate}[(a)]
    \setcounter{enumi}{2}
  \item\label{case4} $v = w$ or $v\in I_3(T)$; or
    \item\label{case5} $v\in D_2(T)$ and there is a good cotree
      edges $l_v v$ with $l_v\not\in \{l, l_u\}$.
    \end{enumerate}
  \end{itemize}
  Moreover, after \eqref{case3} and \eqref{case4}, apply
  Rule~\ref{rule:1} to $l_v v$ and $l_u u$ respectively; after
  \eqref{case5}, apply Rule~\ref{rule:1} to $l_u u$ and then to $l_v
  v$ (if still applicable).
\end{exrule}

One can check in polynomial time whether Rule~\ref{rule:2} (and which
case of it) is applicable.  To show that the whole procedure can be
finished in polynomial time, we need to argue that each invocation
increases the number of internal vertices by at least one.  Note that
Rule~\ref{rule:2} is only applied after Rule~\ref{rule:1} is no longer
applicable.
\begin{lemma}\label{lem0}
  Applying Rule~\ref{rule:1} or Rule~\ref{rule:2} to a spanning tree
  $T$ of $G$ results in a new spanning tree $T'$ of $G$ satisfying
  $|I(T')| \ge |I(T)| + 1$.
\end{lemma}
\begin{proof}
  In Rule~\ref{rule:1} and Rule~\ref{rule:2}, the replaced edge $u v$
  is in $P_T(l_1, l_2)$ and $P_T(l, w)$ respectively, so the resulting
  subgraph $T'$ must be a spanning tree of $G$.  To compare the number
  of internal vertices, it suffices to consider these vertices
  incident to the added/deleted edges.

  After the application of Rule~\ref{rule:1}, $u$ remains an internal
  vertex as $d_{T'}(u) = d_T(u) - 1\ge 2$.  Note that possibly $v\in
  \{l_1, l_2\}$, and in this case $v$ remains a leaf but the other
  leaf becomes an internal vertex.  Otherwise, both $l_1$ and $l_2$
  become internal vertices, while $v$ might become a leaf.  In either
  case, the number of internal vertices increases by at least $1$.

  We now consider Rule~\ref{rule:2}.  Case \eqref{case2} is
  straightforward: after its application, all vertices in $\{l, w, u,
  v\}$ are internal vertices of the resulting tree $T'$; hence
  $|I(T')| = |I(T)| + 1$.  In case \eqref{case3}, after the
  substitution, $w$ and $u$ remain internal vertices, while $l$
  becomes an internal vertex and $v$ becomes a leaf of the new tree.
  Albeit the number of leaves does not change, the cotree edge of the
  new tree between $l_v$ and the new leaf $v$ enables us to apply
  Rule~\ref{rule:1}, which increases the number of internal vertices
  by $1$.  Case \eqref{case4} is similar as case \eqref{case3}: the
  first substitution does not affect the number of internal vertices,
  but the subsequent application of Rule~\ref{rule:1} increases it by
  $1$.  In case \eqref{case5}, after the first substitution, $w$
  remains an internal vertex, $l$ becomes an internal vertex, while
  both $u$ and $v$ become leaves; hence the number of leaves decreases
  by $1$.  Since $l_u$, $u$, $l_v$, $v$ are four distinct leaves in
  the resulting tree, applying Rule~\ref{rule:1} to $l_u u$ results in
  a spanning tree of $G$ that has at least $I(T)$ internal vertices.
  Either $l_v$ and $v$ (which remain leaves) are the only leaves and
  we are done, or we can apply Rule~\ref{rule:1} to $l_v v$ to
  increase the number of internal vertices.  This obtained tree $T'$
  has at least one more internal vertex than $T$.
\end{proof}

A spanning tree $T$ of a graph $G$ is called {\em maximal} if neither
exchange rule is applicable to it.  By Lemma~\ref{lem0}, each
application of an exchange rule to a spanning tree increases its
number of internal vertices at least $1$.  Since a spanning tree of
$G$ has at most $|V(G)| - 2$ internal vertices, we have the following
theorem.
\begin{theorem}\label{thm:maximal-trees}
  A maximal spanning tree of a graph can be constructed in polynomial
  time.
\end{theorem}

\section{The kernelization algorithm}
Although neither exchange rule reduces the graph size, their
exhaustive application provides a maximal spanning tree whose
structural properties will be crucial for our kernelization algorithm.
In this section we are only concerned with maximal spanning trees.  We
will use the reduction rule of Fomin et
al.~\cite{fomin-13-kernel-mist}, which is recalled below.  Let \opt{G}
denote the maximum number of internal vertices a spanning tree of $G$
can have.
\begin{lemma}[\cite{fomin-13-kernel-mist}]
  \label{lem8}
  Let $L'$ be an independent set of $G$ such that $|L'| \ge 2
  |N_G(L')|$.  We can find in polynomial time nonempty subsets
  $S\subseteq N_G(L')$ and $L\subseteq L'$ such that:
  \begin{enumerate}[(1)]
  \item $N_G(L) = S$, and
  \item the graph $(S\cup L, E(G)\cap (S \times L))$ has a spanning
    tree such that all vertices of $S$ and $|S|-1$ vertices of $L$ are
    internal.
  \end{enumerate}
  Moreover, let $G'$ be obtained from $G$ by adding a vertex $v_{\mbox
    {\tiny $S$}}$ adjacent to every vertex in $N_G(S)\backslash L$,
  adding a vertex $v_{\mbox {\tiny $L$}}$ adjacent to $v_{\mbox {\tiny
      $S$}}$, and removing all vertices of $S\cup L$, then $\opt{G'} =
  \opt{G} - 2|S| + 2$.
\end{lemma}
Note that $|L| \ge 2$ (otherwise the graph in Lemma~\ref{lem8}(2)
cannot have internal vertices), and hence each application of the
reduction rule decreases the number of vertices by at least $1$.  The
safeness of the following reduction rule is ensured by
Lemma~\ref{lem8}.

\paragraph{Reduction Rule} (\cite{fomin-13-kernel-mist}): Find
nonempty subsets $S$ and $L$ of vertices as in Lemma~\ref{lem8}.
Return ($G', k'$) where $G'$ is defined in Lemma~\ref{lem8} and $k' =
k - 2|S| + 2$.

\medskip

The main technical obstacle is then to identify a vertex set $L'$ with
$|L'| \ge 2 |N_G(L')|$.  The is trivial when $|V(G)| \ge 3 k - 3$.  In
any spanning tree $T$ of $G$ with $|L(T)| < k$ it holds that $|L(T)|
\ge 2k - 2 \ge 2 |I(T)| = 2 |N_G(L(T))|$; hence we can use $L(T)$ as
$L'$ and a $3 k$-vertex kernel follows.  However, it becomes very
nontrivial to find such a set when $2k < |V(G)| < 3 k - 3$.  Our
approach here is to separate a maximal spanning tree $T$ into several
subtrees and bound the number of $L(T)$ by the number of $I(T)$
residing in each subtree individually.  It is worth mentioning that a
leaf of a subtree may not be a leaf of $T$.

Recall that the removal of any edge $u v\in E(T)$ from a tree $T$
breaks it into two components, one containing $u$ and the other
containing $v$.  In general, the removal of all edges of an edge
subset $E'\subseteq E(T)$ from $T$ breaks it into $|E'| + 1$
components, each being a subtree of $T$.  We would like to divide $T$
in a way that the two ends of any good cotree edge always reside in
the same subtree, hence the following definition.
\begin{definition}
  An edge $u v\in E(T)$ connecting two internal vertices $u,v$ of $T$
  is {\em critical} if there is no good cotree edge connecting the two
  components of $T - uv$.
\end{definition}
Let $C(T)$ denote the (possibly empty) set of all critical edges in
$T$.  Note that for each non-critical edge $u v$ with both $u, v \in
D(T)$, there must be a good cotree edge connecting the two components
of $T - uv$.
\begin{lemma}\label{lem1}
  Let $u$ and $v$ be two detachable vertices of a maximal spanning
  tree $T$ and $u v\in E(T)$.  If there is a good cotree edge
  connecting the two components of $T - uv$, then there exists a good
  cotree edge $l w$ such that
  \begin{enumerate}[(1)]
  \item $u\in D_2(T)$, the only good cotree edge incident to $u$ is $l
    u$, and $l w$ crosses $u v$; or
  \item $v\in D_2(T)$, the only good cotree edge incident to $v$ is $l
    v$, and $l w$ crosses $v u$.
  \end{enumerate}
\end{lemma}
\begin{proof}
  Let $l' w'$ be a good cotree edge connecting the two components of
  $T - uv$.  Without loss of generality, we may assume that $l' w'$
  cross $u v$---then $l'$ and $w'$ are in the components of $T - uv$
  containing $u$ and $v$ respectively---the other case follows by
  symmetry.

  We argue first that $u\in D_2(T)$.  Suppose for contradiction, $u\in
  I_3(T)$.  Since Rule~\ref{rule:2}\eqref{case2} is not applicable to
  $l' w'$ and $u v$, we must have $v\not\in I_3(T)$.  Then $v\in
  D_2(T)$, and by definition, there is a good cotree edge $l_v v$ of
  $T$.  Again, since Rule~\ref{rule:2}\eqref{case2} is not applicable
  to $l_v v$ and $u v$, we must have $l_v\ne l'$.  However, $l_v \neq
  l'$ and $v \neq w$ would imply that Rule~\ref{rule:2}\eqref{case3}
  is applicable to $l' w'$, $u v$, and $l_v v$, a contradiction.  Now
  that $u\in D_2(T)$, by definition, there is a good cotree edge
  incident to $u$.  The proof is now completed if $l' u$ is a good
  cotree edge of $T$ and the only one incident to $u$: we are in case
  (1) and $l' w'$ is the claimed edge.  In the rest of the proof we
  may assume otherwise---that is, there is a good cotree edge $l u$ of
  $T$ with $l \ne l'$.

  Since Rule~\ref{rule:2}\eqref{case4} cannot be applicable to $l'
  w'$, $u v$, and $l u$, the vertex $v$ is also in $D_2(T)$ and $v
  \neq w'$.  There is also a good cotree edge $l_v v$ of $T$.
  Similarly it can be inferred that $l_v\ne l'$: otherwise
  Rule~\ref{rule:2}\eqref{case4} is applicable to $l' v$ (i.e., $l_v
  v$), $u v$, and $l u$.  Now that $l'$ is different from both $l$ and
  $l_v$ but Rule~\ref{rule:2}\eqref{case5} is not applicable to $l'
  w'$, $u v$, $l u$ and $l_v v$, we must have $l = l_v$, i.e., both $l
  u$ and $l v$ are good cotree edges of $T$.  We consider now which
  component of $T - uv$ the leaf $l$ belongs to.  If it is with $u$,
  then there cannot be another good cotree edge $l'_u u$ with $l'_u
  \neq l$: otherwise, Rule~\ref{rule:2}\eqref{case4} would be
  applicable on $l v$, $u v$, and $l'_u u$.  In other words, $l u$ is
  the only good cotree edge incident to $u$, and $l v$ is the claimed
  good cotree edge crossing $u v$.  We are in case (1) and $l v$ is
  the claimed edge.  A symmetric argument implies that we are in case
  (2) and $l u$ is the claimed edge if $l$ is in the component of $T -
  uv$ with $v$.
  This concludes the proof.
\end{proof}
The vertices $u$ or $v$ stipulated in Lemma \ref{lem1} turns out to be
our main trouble in analyzing the size of reduced instance.  We use
$D_B(T)$ to denote this set of vertices, which need our special
attention.  For the pair of vertices $u,v$ as in Lemma~\ref{lem1},
$u\in D_B(T)$ if case (1) holds true, and $v\in D_B(T)$ otherwise.
Note that $D_B(T) \subseteq D_2(T) \subseteq I_2(T)$.  This vertex has
degree $2$ in $T$, and its other neighbor (different from $v$ or $u$)
cannot be a leaf of $T$: suppose that it is $u$; the component of $T -
uv$ containing $u$ must contain another $l\in L(T)$ that is
nonadjacent to $u$ in $T$.  Therefore, the only good cotree edge
mentioned in Lemma \ref{lem1} is actually the only edge between it and
$L(T)$ in $G$.  The following corollary follows easily.
\begin{corollary}\label{cor1.5}
  For each $u\in D_B(T)$ for a maximal spanning tree $T$, we have
  $|L(T)\cap N_G(u)| = 1$.
\end{corollary}
Note that $I_3(T) \subseteq D(T)\setminus D_B(T)$ and $I(T)\setminus
D(T)\subseteq I_2(T)$.

\begin{lemma}\label{lem6}
  Let $T$ be a maximal spanning tree.  For every pair of vertices $u,
  w\in D(T)\setminus D_B(T)$ that are in the same component of $T -
  C(T)$, the path $P_T(u, w)$ visits at least one vertex $v\in
  I(T)\setminus D(T)$ such that for any $l \in L(T)$, the path $P_T(l,
  v)$ visits $D(T)$.
\end{lemma}
\begin{proof}
  No generality will be lost by assuming that $P_T(u, w)$ is minimal
  (in the sense that it visits no other vertex in $D(T)\setminus
  D_B(T)$).  By assumption, $P_T(u, w)$ is retained in $T - C(T)$.  We
  argue first $u w\not\in E(T)$.  Suppose for contradiction, $u w\in
  E(T)$.  Since $u v\not\in C(T)$, there must be some good cotree edge
  crossing it; however, by Lemma~\ref{lem1} and the definition of
  $D_B(T)$, at least one of $u$ and $w$ is then in $D_B(T)$, a
  contradiction.  Let $P_T(u, w) = u v_1 \cdots v_p w$, where $p \ge
  1$; note that $d_T(v_i) = 2$ and $v_i\not\in D(T)\setminus D_B(T)$
  for each $1\le i\le p$.

  We now find an internal vertex of $P_T(u, w)$ that is not in $D(T)$
  as follows.  If $v_1\not\in D(T)$, then we are done.  Otherwise,
  $v_1\in D_B(T)$, and there is a unique good cotree edge $l v_1$.  We
  prove by contradiction that $l$ is in the same component of $T - u
  v_1$ with $v_1$.  Suppose the contrary, then
  \begin{itemize}
  \item if $u\in I_3(T)$, then Rule~\ref{rule:2}\eqref{case2} is
    applicable to $l v_1$ and $u v_1$; or
  \item if $u\in D_2(T)\setminus D_B(T)$, then there is a good cotree
    edge $l' u$ with $l'\neq l$, and hence
    Rule~\ref{rule:2}\eqref{case4} is applicable to $l v_1$, $u v_1$,
    and $l' u$.
  \end{itemize}
  Noting that every internal vertex of $P$ has degree $2$, this
  actually implies that $l$ must be in the same component of $T - v_p
  w$ with $w$.  The first $v_i$ such that $l v_i$ is {\em not} the
  only good cotree edge incident to $v_i$ is the vertex we need.  Its
  existence can be argued by contradiction as follows.  Suppose for
  contradiction, there is no such a vertex $v_i$, then $l v_p$ is the
  only good cotree edge incident to $v_p$, and
  \begin{itemize}
  \item if $w\in I_3(T)$, then Rule~\ref{rule:2}\eqref{case2} is
    applicable to $l v_p$ and $w v_p$; or
  \item if $w\in D_2(T)\setminus D_B(T)$, then there is a good cotree
    edge $l' w$ with $l'\neq l_1$, and hence
    Rule~\ref{rule:2}\eqref{case4} is applicable to $l v_p$, $w
    v_p$, and $l' w$.
  \end{itemize}
  These contradictions imply that $p \ge 2$ and there must be $2\le
  i\le p$ such that $v_i$ is incident to a good cotree edge $l' v_i$
  with $l' \ne l$.  Since Rule~\ref{rule:2}\eqref{case4} is not
  applicable to $l v_{i-1}$, $v_i v_{i-1}$, and $l' v_i$, we can
  conclude that $v_i\not\in D_2(T)$.  Noting that $v_i\in I_2(T)$, we
  have verified that $v_i$ is an internal vertex of $P_T(u,w)$ not in
  $D(T)$.

  Noting that every internal vertex of $P_T(u,w)$ has degree $2$, for
  any $l \in L(T)$, the path $P_T(l,v)$ necessarily visits either $u$
  or $w$, which is in $D(T)$.  This concludes the proof.
\end{proof}

We are now ready for proving the main result of this section.
\begin{lemma}\label{lem:reduction}
  Let $T$ be a maximal spanning tree of $G$ with $|V(G)|\ge 4$.  If
  $|L(T)| > |I(T)|$, then we can find in polynomial time an
  independent set $L'$ of $G$ such that $|L'| \ge 2|N_G(L')|$.
\end{lemma}
\begin{proof}
  We find all critical edges $C(T)$, and take the forest $T - C(T)$.
  By assumption, there must be some component $T_0$ of $T - C(T)$ of
  which more than half vertices are from $L(T)$.  Let $X$ and $Y$
  denote $L(T)\cap V(T_0)$ and $I(T)\cap V(T_0)$ respectively; then
  $|X| \ge |Y| + 1$.  Since $T$ is maximal (Rule~\ref{rule:1} is not
  applicable), $X$ is an independent set.  We divide $X$ into the
  following three subsets:
  \[
  X_1 := X\cap N_G(D_B(T));\qquad X_2 := X\cap N_T(D(T)) \setminus X_1;
  \quad \text{ and }\quad X_3 := X\setminus (X_1\cup X_2).
  \]
  We will show that $|X_2| \ge 2 |N_G(X_2)|$, and hence $X_2$
  satisfies the claimed condition and can be used as $L'$.  By the
  definition of critical edges, there is no good cotree edge of $T$
  connecting two different components of $T - C(T)$; hence
  $N_G(X)\subseteq Y$.  We accordingly divide $Y$ into subsets.  The
  detachable vertices are either in $Y_1 := D_B(T)\cap V(T_0)$ or $Y_2
  := \left( D(T)\setminus D_B(T) \right) \cap V(T_0)$, while a vertex
  $y\in Y\setminus D(T)$ is in $Y_3$ if there exists $l\in L(T)$ such
  that the path $P_T(l,y)$ does not visit $D(T)$, or in $Y_4$
  otherwise.  Note that $|X| = |X_1| + |X_2| + |X_3|$ and $|Y| =
  |Y_1| + |Y_2| + |Y_3| + |Y_4|$.

  We argue first that $N_G(X_2)\subseteq Y_2$.  It suffices to show
  $N_G(X_2)\subseteq D(T)$ (the definition of $X_2$ requires that a
  vertex in it is nonadjacent to $D_B(T)$ in $G$), which further boils
  down to showing $N_G(X_2)\cap I_2(T)\subseteq D(T)$: since $T$ is
  maximal (Rule~\ref{rule:1} is not applicable), $X_2$ has no neighbor
  in $L(T)$; on the other hand, $I_3(T)\subseteq D(T)$.  Consider a
  vertex $x \in X_2$, and let $y$ be the unique neighbor of $x$ in
  $T$.  By assumption, $y\in D_2(T)\setminus D_B(T)$, and hence there
  is a good cotree edge $l y$ of $T$ with $l\neq x$.  For each $y'\in
  N_G(x)\cap I_2(T)$ different from $y$, the path $P_T(x, y')$ visits
  $y$, using the definition of $D_2(T)$ we can conclude that $y'\in
  D_2(T)$.

  Each $x\in X_1$ has a neighbor $y \in Y_1$.  By
  Corollary~\ref{cor1.5}, $x$ is the only vertex in $N_G(y) \cap
  L(T)$.  Thus, $|X_1| \le |Y_1|$.  The unique neighbor $y$ of a
  vertex $x\in X_3$ in $T$ must be in $I_2(T)\setminus D_2(T)$.  Since
  the trivial path $P_T(x, y)$ (consisting of a single edge $xy$) does
  not visit $D(T)$, we have $y\in Y_3$.  The other neighbor of $y$ in
  $T$ cannot be a leaf of $T$ ($G$ has at least four vertices).  Thus,
  $|X_3| \le |Y_3|$.  By Lemma~\ref{lem6}, for any two different
  vertices $u$ and $w$ of $Y_2$, the path $P_T(u,w)$ visits at least
  one vertex in $Y_4$.  Since $T_0$ is a tree, using induction it is
  easy to show $|Y_4| \ge |Y_2| - 1$.

  Summarizing above, we have
  \begin{align*}
    |X_2| &= |X| - |X_1| - |X_3| & \qquad (\text{because }|X| = |X_1|
    + |X_2| + |X_3|.)
    \\
    &\ge |Y| + 1 - |Y_1| - |Y_3| & \qquad (\text{because }|X| \ge |Y|
    + 1; |X_1| \le |Y_1|; |X_3| \le |Y_3|.)
    \\
    &= |Y_2| + |Y_4| + 1 & \qquad (\text{because }|Y| = |Y_1| + |Y_2|
    + |Y_3| + |Y_4|.)
    \\
    &\ge 2|Y_2| & \qquad (\text{because }|Y_4| \ge |Y_2| - 1.)
    \\
    &\ge 2|N_G(X_2)|.& \qquad (\text{because }N_G(X_2)\subseteq Y_2.)
  \end{align*}
  Hence $X_2$ can be used as the independent set $L'$.  This concludes
  the proof.
\end{proof}
Lemmas~\ref{lem:reduction} and \ref{lem8}, together with
Theorem~\ref{thm:maximal-trees}, imply Theorem~\ref{thm:main}.

redpajama_set_name: RedPajamaArXiv


text:
Q: Steinhaus Theorem proof on Wikipedia I would like help understanding this proof of the Steinhaus Theorem.
I have two questions:

*

*In the proof, they claim that there exists an interval $\Delta =(a,b)$ containing E so that $m(E)>\alpha (b-a)$. Can someone justify the validity of this claim?

*In the proof, they allow $m(E)$ to be infinite. Why doesn't the proof break-down in this case?

I've already thought a bit about this:
I know that there exists an open set $U$ containing $E$ so that $m(U)<m(E)+\varepsilon$. But, in the notation of the proof linked above, how is the interval $\Delta$ found? Is it some union relating to $U$?